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3 years ago

Object a has a mass of 20 g

Physics

1 answer:

Anarel [89]3 years ago

7 0

And.. where is the rest of the question?

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You have a meteorite sample and you decide to use the uranium-235/lead-207 system to date it. After analysis, you find that it h

storchak [24]

Originally there must been

1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start

% = 2.25 / 150 = 1.5 % of 235 U left

5 0

2 years ago

26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the

denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

$U=mg \Delta h$

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h = 4 m + 0.02 m = 4.02 m$ is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

$\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J$

4 0

3 years ago

Objects in free fall are weightless. true or false???

frez [133]

The answer is false

7 0

3 years ago

Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving

Iteru [2.4K]

Explanation:

Let $m_p$ is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

$\dfrac{mv^2}{r}=qvB$

r is the radius of path,

$r=\dfrac{mv}{qB}$

Time period is given by :

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi m_p}{qB}$

Frequency of proton is given by :

$f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}$

The wavelength of radiation is given by :

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{2\pi m_pc}{qB}$

So, the wavelength of radiation produced by a proton is $\dfrac{2\pi m_pc}{qB}$ . Hence, this is the required solution.

3 0

3 years ago

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

4 0

3 years ago